hdu4490 Mad Veterinarian(bfs)

Mad Veterinarian

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 249    Accepted Submission(s): 104
Special Judge

Problem Description

Mad Veterinarian puzzles have a mad veterinarian, who has developed several machines that can transform an animal into one or more animals and back again. The puzzle is then to determine if it is possible to change one collection of animals into another by applying the machines in some order (forward or reverse). For example:

Machine A turns one ant into one beaver.

Machine B turns one beaver into one ant, one beaver and one cougar.

Machine C turns one cougar into one ant and one beaver.

Can we convert a beaver and a cougar into 3 ants?

Can we convert one ant into 2 ants? NO

These puzzles have the properties that:

1. In forward mode, each machine converts one animal of a given species into a finite, non-empty collection of animals from the species in the puzzle.

2. Each machine can operate in reverse.

3. There is one machine for each species in the puzzle and that machine (in forward mode) takes as input one animal of that species.

Write a program to find the shortest solution (if any) to Mad Veterinarian puzzles. For this problem we will restrict to Mad Veterinarian puzzles with exactly three machines, A, B, C.

 

Input

The first line of input contains a single integer P, (1<= P <= 1000 ), which is the number of data sets that follow. Each data set consists of several lines of input. Each data set should be processed identically and independently.

The first line of each data set consists of two decimal integers separated by a single space. The first integer is the data set number. The second integer is the number, N, of puzzle questions. The next three input lines contain the descriptions of machines A, B and C in that order. Each machine description line consists of three decimal integers separated by spaces giving the number of animals of type a, b and c output for one input animal. The following N lines give the puzzle questions for the Mad Veterinarian puzzle. Each contains seven decimal digits separated by single spaces: the puzzle number, the three starting animal counts for animals a, b and c followed by the three desired ending animal counts for animals a, b and c.

 

Output

For each input data set there are multiple lines of output. The first line of output for each data set contains the data set number, a space and the number of puzzle questions (N). For each puzzle question, there is one line of output which consists of the puzzle question number followed by a space, followed by “NO SOLUTION”, (without the quotes) if there is no solution OR the puzzle question number followed by the shortest number of machine steps used, a space and a sequence of letters [A B C a b c] with capital letters indicating applying the machine in the forward direction and lower case letters indicating applying the machine in the reverse direction.

 

Sample Input

2
1 2
0 1 0
1 1 1
1 1 0
1 0 1 1 3 0 0
2 1 0 0 2 0 0
2 2
0 3 4
0 0 5
0 0 3
1 2 0 0 0 0 5
2 2 0 0 0 0 4

 

Sample Output

1 2
1 3 Caa
2 NO SOLUTION
2 2
1 NO SOLUTION
2 25 AcBcccBccBcccAccBccBcccBc

 

Source

Greater New York 2012

 

题意：有三个机器ABC,可以分别把a,b,c变成对应的某几个东西，也可以逆着变，求变到目标状态的最少步数

很水的题，不过我写了好长好长，太不符合我的风格了，刚开始调试的时候无解的情况表示不出来，因为它的数据可以一直往上涨，最后限制了一下，看学长的博客说的最多8个，加这个限制就可以了，好不容易调试出来了，结果跟示例对不上，过了好久才想到方法可能不止一种，于是提交了，结果就AC了。。。0ms

写的有点长，不过很多就是差不多的，看起来应该不费力吧，这题的输入数据很蛋疼要小心

#include<stdio.h>
#include<string.h>
#define M 8
struct node
{
	int a,b,c,f;
	char o;
}f[4],s[1000],start,end;
int visit[10][10][10];
int bfs()
{
	int t=0,w=1;
	memset(visit,0,sizeof(visit));
	visit[s[t].a][s[t].b][s[t].c]=1;
	s[0].f=-1;
	while(t<w)
	{
		if(s[t].a>0)
		{
			s[w].a=s[t].a+f[0].a-1;
			s[w].b=s[t].b+f[0].b;
			s[w].c=s[t].c+f[0].c;
			s[w].f=t;
			s[w].o='A';
			if(s[w].a<M&&s[w].b<M&&s[w].c<M&&!visit[s[w].a][s[w].b][s[w].c])
			{
				if(s[w].a==end.a&&s[w].b==end.b&&s[w].c==end.c)
				{
					end.f=w;
					return w;
				}
				visit[s[w].a][s[w].b][s[w].c]=1;
				w++;
			}
		}
		if(s[t].b>0)
		{
			s[w].b=s[t].b+f[1].b-1;
			s[w].a=s[t].a+f[1].a;
			s[w].c=s[t].c+f[1].c;
			s[w].f=t;
			s[w].o='B';
			if(s[w].a<M&&s[w].b<M&&s[w].c<M&&!visit[s[w].a][s[w].b][s[w].c])
			{
				if(s[w].a==end.a&&s[w].b==end.b&&s[w].c==end.c)
				{
					end.f=w;
					return w;
				}
				visit[s[w].a][s[w].b][s[w].c]=1;
				w++;
			}
		}
		if(s[t].c>0)
		{
			s[w].c=s[t].c+f[2].c-1;
			s[w].b=s[t].b+f[2].b;
			s[w].a=s[t].a+f[2].a;
			s[w].f=t;
			s[w].o='C';
			if(s[w].a<M&&s[w].b<M&&s[w].c<M&&!visit[s[w].a][s[w].b][s[w].c])
			{
				if(s[w].a==end.a&&s[w].b==end.b&&s[w].c==end.c)
				{
					end.f=w;
					return w;
				}
				visit[s[w].a][s[w].b][s[w].c]=1;
				w++;
			}
		}
		if(s[t].a>=f[0].a&&s[t].b>=f[0].b&&s[t].c>=f[0].c)
		{
			s[w].a=s[t].a-f[0].a+1;
			s[w].b=s[t].b-f[0].b;
			s[w].c=s[t].c-f[0].c;
			s[w].f=t;
			s[w].o='a';
			if(s[w].a<M&&s[w].b<M&&s[w].c<M&&!visit[s[w].a][s[w].b][s[w].c])
			{
				if(s[w].a==end.a&&s[w].b==end.b&&s[w].c==end.c)
				{
					end.f=w;
					return w;
				}
				visit[s[w].a][s[w].b][s[w].c]=1;
				w++;
			}
		}
		if(s[t].a>=f[1].a&&s[t].b>=f[1].b&&s[t].c>=f[1].c)
		{
			s[w].b=s[t].b-f[1].b+1;
			s[w].a=s[t].a-f[1].a;
			s[w].c=s[t].c-f[1].c;
			s[w].f=t;
			s[w].o='b';
			if(s[w].a<M&&s[w].b<M&&s[w].c<M&&!visit[s[w].a][s[w].b][s[w].c])
			{
				if(s[w].a==end.a&&s[w].b==end.b&&s[w].c==end.c)
				{
					end.f=w;
					return w;
				}
				visit[s[w].a][s[w].b][s[w].c]=1;
				w++;
			}
		}
		if(s[t].a>=f[2].a&&s[t].b>=f[2].b&&s[t].c>=f[2].c)
		{
			s[w].c=s[t].c-f[2].c+1;
			s[w].b=s[t].b-f[2].b;
			s[w].a=s[t].a-f[2].a;
			s[w].f=t;
			s[w].o='c';
			if(s[w].a<M&&s[w].b<M&&s[w].c<M&&!visit[s[w].a][s[w].b][s[w].c])
			{
				if(s[w].a==end.a&&s[w].b==end.b&&s[w].c==end.c)
				{
					end.f=w;
					return w;
				}
				visit[s[w].a][s[w].b][s[w].c]=1;
				w++;
			}
		}
		t++;
	}
	return 0;
}
int main()
{
	int t,i,j,n,num,tt;
	char cou[100];
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&num,&n);
		printf("%d %d\n",num,n);
		for(i=0;i<3;i++)
			scanf("%d%d%d",&f[i].a,&f[i].b,&f[i].c);
		while(n--)
		{
			scanf("%d%d%d%d%d%d%d",&tt,&s[0].a,&s[0].b,&s[0].c,&end.a,&end.b,&end.c);
			printf("%d ",tt);
			if(i=bfs())
			{
				for(j=98;i>0;j--)
				{
					cou[j]=s[i].o;
					i=s[i].f;
				}
				cou[99]='\0';
				printf("%d ",98-j);
				puts(cou+j+1);
			}
			else puts("NO SOLUTION");
		}
	}
	return 0;
}